JEE Main 2026 — Physics Electromagnetism

The magnetic flux linked with the circular loop is given by $\Phi = BA \cos \theta$Φ=BAcosθ. Since the plane of the loop is perpendicular to the magnetic field, the angle between the area vector and the magnetic field is $\theta = 0^{\circ}$θ=0∘.

$$\Phi = B \cdot \pi r^2$$Φ=B⋅πr2

Given $r = 20\text{ cm} = 0.2\text{ m}$r=20 cm=0.2 m, the area of the loop is:

$$A = \pi (0.2)^2 = 0.04\pi\text{ m}^2$$A=π(0.2)2=0.04π m2

The induced emf in the loop is given by Faraday's law of induction:

$$|E| = \dfrac{d\Phi}{dt} = A \dfrac{dB}{dt}$$∣E∣=dtdΦ​=AdtdB​

Given the magnetic field $B = 2t^2 + 2t + 3$B=2t2+2t+3, we differentiate it with respect to time $t$t:

$$\dfrac{dB}{dt} = \dfrac{d}{dt}(2t^2 + 2t + 3) = 4t + 2$$dtdB​=dtd​(2t2+2t+3)=4t+2

At $t = 3\text{ s}$t=3 s, the rate of change of the magnetic field is:

$$\dfrac{dB}{dt} = 4(3) + 2 = 14\text{ T/s}$$dtdB​=4(3)+2=14 T/s

Now, substituting the values into the emf equation:

$$|E| = 0.04\pi \times 14 = 0.56\pi\text{ V}$$∣E∣=0.04π×14=0.56π V

The induced current $I$I in the loop is:

$$I = \dfrac{|E|}{R} = \dfrac{0.56\pi}{2} = 0.28\pi\text{ A}$$I=R∣E∣​=20.56π​=0.28π A

Using $\pi = \dfrac{22}{7}$π=722​, we get:

$$I = 0.28 \times \dfrac{22}{7} = 0.04 \times 22 = 0.88\text{ A}$$I=0.28×722​=0.04×22=0.88 A

We are given that the induced current is $\dfrac{\alpha}{50}\text{ A}$50α​ A. Equating the two values:

$$\dfrac{\alpha}{50} = 0.88$$50α​=0.88

$$\alpha = 0.88 \times 50 = 44$$α=0.88×50=44

Answer: $44$44